# Why do some of the Feynman propagators for a complex scalar field vanish?

I am learning QFT. Earlier we showed that a complex field can be decomposed like so: \begin{align*} \phi &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\big(a(\mathbf{k})e^{-ikx}+b^\dagger(\mathbf{k})e^{ikx}\big) \\ \phi^\dagger &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\big(b(\mathbf{k})e^{-ikx}+a^\dagger(\mathbf{k})e^{ikx}\big). \end{align*} An exercise was given to me to show that the Feynman propagators $$\langle 0 | T \phi(x)\phi(y) |0 \rangle$$ and $$\langle 0 | T \phi^\dagger(x)\phi^\dagger(y) |0 \rangle$$ are both zero. Unfortunately, I'm not getting zero in my calculation, so what am I doing wrong?

This is what I have. First, \begin{align*} \phi(y)|0\rangle &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}} b^\dagger(\mathbf{k})e^{iky} |0\rangle \\ &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} e^{iky} |\mathbf{k} \rangle . \\ \langle 0 | \phi(x) & = \int \frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}} \langle 0| a(\mathbf{k})e^{-ikx} \\ &= \int \frac{d^3k}{(2\pi)^3}\frac{1}{2\omega_k} \langle \mathbf{k} | e^{-ikx} .\end{align*} Then combining them, using $$\langle \mathbf{k} | \mathbf{k'} \rangle = (2\pi)^3 2\omega_k \delta(\mathbf{k}-\mathbf{k}')$$, and integrating over one of the $$k$$'s, $$\langle 0 | T \phi(x)\phi(y) |0 \rangle = \int \frac{d^3k}{(2\pi)^3} \frac{1}{2\omega_k}e^{ik(y-x)}$$ assuming that $$x^0>y^0$$. But if $$x^0 < y^0$$, then it seems we can just switch $$y$$ and $$x$$ (unless this is where the issue lies?).

I found a few sites saying something like that $$\phi$$ creates a particle but destroys an antiparticle, or was it the other way around? In any case, it is not clear to me how that is the case at all. Isn't $$b^\dagger$$ the creation operator for particles and $$b$$ the annihilation operator for antiparticles? I also fail to see how this answers why the propagator vanish.

Note, that $$a^{\dagger} (\mathbf k)$$ and $$b^{\dagger} (\mathbf k)$$ create particles of different kind, so they commute with each other: $$[a(\mathbf k), b^{\dagger} (\mathbf k^{'}) ] = [a^{\dagger}(\mathbf k), b (\mathbf k^{'}) ] = 0 \qquad$$ The simply use: $$a (\mathbf k) | 0 \rangle = 0, \quad \langle 0 | b^{\dagger} (\mathbf k) = 0$$. In the cross term, due to the afomentioned property, one may move raising and lowering operators, such that they annihilate vacuum states.